A projectile is launched from ground level with an initial velocity of 50 m/s at an angle of 30° above the horizontal. Taking g = 10 m/s^2 and neglecting air resistance, what is the total flight time until the projectile returns to ground level?
Answer and explanation
Correct answer: A. 5.0 seconds
- Step 1: Calculate the initial vertical velocity component: v_0y = v_0 sin(30°) = 50 0.5 = 25 m/s.
- Step 2: At peak height, vertical velocity is 0: v_y = v_0y - g * t_peak = 0 => t_peak = 25 / 10 = 2.5 s.
- Step 3: Total flight time is double the time to peak: t_total = 2 t_peak = 2 2.5 = 5.0 s.
Answer
5.0 seconds.
Why not the others: '2.5 seconds' is the time to reach maximum height; '10.0 seconds' doubles total flight time; '4.3 seconds' uses the horizontal velocity component.
Practise more Classical Mechanics and Electromagnetism.
Free questions with an explanation for every one, and full timed Qatar Senior Secondary Certificate (Thanaweya) mock tests.
More Classical Mechanics and Electromagnetism questions
- A 1200 kg car rounds an unbanked circular curve of radius 50 meters on a horizontal road. If the coefficient of static friction between the…
- A constant force of 40 N acts on a 5.0 kg body initially at rest for a duration of 3.0 seconds. What is the work done on the body by this fo…
- A simple pendulum has a period of 2.0 seconds at Earth's surface. What would be its period if it were placed on the Moon, where the gravitat…
- Two parallel plates separated by a distance of 5.0 mm have a potential difference of 250 V across them. What is the magnitude of the uniform…
- An electric circuit consists of a 12 V battery connected across two parallel resistors of 6 ohms and 12 ohms. What is the total current draw…
- A straight conductor of length 0.40 m carrying a current of 5.0 A is positioned in a uniform magnetic field of 0.60 T perpendicular to the f…